How to integrate $\int \sqrt{x^2+a^2}dx$ $a$ is a parameter. I have no idea where to start
 A: Since $\sinh^2(x)+1
=\cosh^2(x)
$,
this suggests
letting $x = a \sinh(y)$.
$dx = a \cosh(y)$
and $x^2+a^2 = a^2(\sinh^2(y)+1)
= a^2 \cosh^2(y)$,
so $a \cosh(y) = \sqrt{x^2+a^2}$.
$\begin{align}
\int{\sqrt{x^2+a^2}}dx
&=\int a^2 \cosh^2(y) dy\\
&= a^2 \int (e^{2y}+2+e^{-2y})\,dy/4\\
&= (a^2/4) (e^{2y}/2 + 2y - e^{-2y}/2)\\
&=(a^2/4)(\sinh(2y)+2y)\\
&=(a^2/4)(2\sinh(y)\cosh(y)+2y)\\
&=(a^2/2)((x/a) (1/a)\sqrt{x^2+a^2}+\sinh^{-1}(x/a))\\
&=(x \sqrt{x^2+a^2})/2+(a^2/2)\sinh^{-1}(x/a)
\end{align}
$
I'll leave it at this - you can find $\sinh^{-1}$.
A: I will give you a proof of how they can get the formula above. As a heads up, it is quite difficult and long, so most people use the formula usually written in the back of the text, but I was able to prove it so here goes.
The idea is to, of course, do trig-substitution.
Since $$\sqrt{a^2+x^2} $$ suggests that $x=a\tan(\theta)$ would be a good one because the expression simplifies to $$a\sec(\theta)$$
We can also observe that $dx$ will become $\sec^2(\theta)d\theta$.
Therefore$$\int \sqrt{a^2+x^2}dx = a^2\int \sec^3(\theta)d\theta$$
Now there are two big things that we are going to do.
One is to do integration by parts to simplify this expression so that it looks a little better, and later we need to be able to integrate $\int \sec(\theta)d\theta$.
So the first step is this. It is well known and natural to let $u=\sec(\theta)$ and $dv=\sec^2(\theta)d\theta$ because the latter integrates to simply, $\tan(\theta)$.
Letting $A = \int \sec^3(\theta)d\theta$,you will get the following $$A = \sec(\theta)\tan(\theta) - \int{\sec(\theta)\tan^2(\theta)d\theta}$$
$$=\sec(\theta)\tan(\theta) - \int{\sec(\theta)d\theta - \int\sec^3(\theta)d\theta}$$
therefore,$$2A = \sec(\theta)\tan(\theta)-\int \sec(\theta)d\theta$$
Dividing both sides give you $$A = 1/2[\sec(\theta)\tan(\theta)-\int \sec(\theta)d\theta]$$
I hope you see now why all we need to be able to do is to integrate $\sec(\theta)$.
The chance that you know how is rather high because you are solving this particular problem, but let's just go through it for the hell of it.
This is a very common trick in integration using trig, but remember the fact that $\sec^2(\theta)$ and $\sec(\theta)\tan(\theta)$ are derivatives of $\tan(\theta)$ and $\sec(\theta)$, respectively. So this is what we do.
$$\int \sec(\theta)d\theta = \int {{\sec(\theta)(\sec(\theta)+\tan(\theta))} \over {\sec(\theta)+\tan(\theta)}} d\theta$$
Letting $w = \sec(\theta)+\tan(\theta)$, 
$$= \int {dw \over w} = \ln|w|$$
So, long story short, 
$$\int \sqrt{a^2+x^2}dx = a^2/2[\sec(\theta)\tan(\theta) - \ln|\sec(\theta)+\tan(\theta)|]$$
$$= {x\sqrt{a^2+x^2}\over 2} + {{a^2\ln|x+\sqrt{a^2+x^2}|}\over 2} + C$$
A: Integrating by parts 
$$I=\int \sqrt{x^2+a^2}\cdot1dx$$
$$=\sqrt{x^2+a^2}\int dx-\int\left( \frac{d \sqrt{x^2+a^2}}{dx}\int dx\right)dx$$
$$=x\sqrt{x^2+a^2}-\int\frac{x^2 dx}{\sqrt{x^2+a^2}}$$
Now, $$\int\frac{x^2 dx}{\sqrt{x^2+a^2}}=\frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx=I-a^2\int\frac{dx}{\sqrt{x^2+a^2}}$$
Put $x=a\tan \theta $ in 
$$\int\frac{dx}{\sqrt{x^2+a^2}}$$
A: Substitue $x=a \sinh t$, so $t=\text{arcsinh} \frac{x}{a} $ $dx=a\cosh t dt$ and $x^2+a^2=a^2(1+\sinh^2t)$. Your integral becomes $$\int|a|\cosh t a\cosh t dt=a|a|\int \cosh^2tdt.$$
Use the identity $\cosh^2t=\frac{1}{2}(1+\cosh2t)$ to get $$\frac{a|a|}{2} \int(1+\cosh2t)dt=\frac{a|a|}{2} \left(t+\frac{\sinh 2t}{2} \right)+C $$
If we go back to $x$, we get: $$\frac{a|a|}{2} \left(\text{arcsinh }\frac{x}{a}+\frac{\sinh2 \text{arcsinh } \frac{x}{a} }{2} \right) .$$
This can be further simplified using the identities $\sinh 2t=2\sinh t \cosh t$ and $\cosh \text{arcsinh t}=\sqrt{1+t^2}$
